Question: Evaluate the improper integral if it exists. $\int^{1}_{-\infty}-e^{x}\,dx $ Choose 1 answer: Choose 1 answer: (Choice A) A $-e$ (Choice B) B $-1$ (Choice C) C $0$ (Choice D) D The improper integral diverges.
Explanation: First, let's rewrite the improper integral: $\int^{1}_{-\infty}-e^{x}\,dx =\lim_{a\to-\infty}\int_a^1 -e^{x}\,dx$ We can now evaluate the integral: $\begin{aligned} \phantom{\int_{-\infty}^{1}e^x\,dx}&=\lim_{a\to-\infty}\int_a^1 -e^{x}\,dx\\ \\ \\ &=-\lim_{a\to-\infty}\int_a^1e^{x}\,dx\\ \\ \\ &=-\lim_{a\to-\infty}\Big[e^x\Big]_a^1\\ \\ \\ &=-\lim_{a\to-\infty}(e^1-e^{a})\\ \\ &=-(\lim_{a\to-\infty}e-\lim_{a\to-\infty}e^a)\\ \\ &=-(e-0)\\ \\ &=-e \end{aligned}$ The answer: $\int^{1}_{-\infty}-e^{x}\,dx =-e$